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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: JackSonSmith on January 02, 2015, 11:40:23 pm

Title: Dificult composite function question.
Post by: JackSonSmith on January 02, 2015, 11:40:23 pm: Let a be a positive number, let f :[2,∞) → R, f (x) = a − x and let g:(−∞,1] → R,
g(x) = x 2 + a. Find all values of a for which f ◦ g and g ◦ f both exist.

Answer is a ∈ [2,3].

Could someone please show me how to solve this, thanks.
Title: Re: Dificult composite function question.
Post by: keltingmeith on January 03, 2015, 12:06:40 am: The easiest way to solve any maths problem like this is to consider all the little cases, and then find the point that all these cases have in common.

So, for fog to exist, we require that $ran_g\subseteq dom_f$ . The domain of f is easy - so, now we just need to find the range of g.

g is a parabola shifted up one unit - and so, the range of it is [a, infinity). This is a subset of [2, infinity) when $a\geq 2$ . So, for fog to exist, we can take $a\geq 2$ .

Now, for gof. This time, we need that $ran_f\subseteq dom_g$ . f is a linear function, and is strictly decreasing. So, its range is going to be (-infinity, a-2]. Now, for this to be a subset of the domain of g, we once again consider the two intervals directly - in this case, the domain of g being (-infinity, 1]. So, we take a-2=1 ===> a=3, so we see that the range of f is a subset of g for $a\leq 3$ .

Finally, we take our two separate cases, and combine them to find the values of a that works for both. So, when is a both greater than 2 AND less than 3 (including those values)? Obviously, when $a\in [2, 3]$ .
Title: Re: Dificult composite function question.
Post by: JackSonSmith on January 03, 2015, 12:46:12 am: Thanks again. I would be so helpless without your help. You would make a great maths teacher.
Title: Re: Dificult composite function question.
Post by: Zues on January 03, 2015, 10:37:13 am: Quote from: EulerFan101 on January 03, 2015, 12:06:40 am
The easiest way to solve any maths problem like this is to consider all the little cases, and then find the point that all these cases have in common.

So, for fog to exist, we require that $ran_g\subseteq dom_f$ . The domain of f is easy - so, now we just need to find the range of g.

g is a parabola shifted up one unit - and so, the range of it is [a, infinity). This is a subset of [2, infinity) when $a\geq 2$ . So, for fog to exist, we can take $a\geq 2$ .

Now, for gof. This time, we need that $ran_f\subseteq dom_g$ . f is a linear function, and is strictly decreasing. So, its range is going to be (-infinity, a-2]. Now, for this to be a subset of the domain of g, we once again consider the two intervals directly - in this case, the domain of g being (-infinity, 1]. So, we take a-2=1 ===> a=3, so we see that the range of f is a subset of g for $a\leq 3$ .

Finally, we take our two separate cases, and combine them to find the values of a that works for both. So, when is a both greater than 2 AND less than 3 (including those values)? Obviously, when $a\in [2, 3]$ .

Hey Euler, how did you know you had to do a-2?
Title: Re: Dificult composite function question.
Post by: keltingmeith on January 03, 2015, 12:45:21 pm: Quote from: Zues on January 03, 2015, 10:37:13 am
Hey Euler, how did you know you had to do a-2?

Because f(x)=a-x, we can use the fact that it's a linear function to find the range by simply finding the values of x at the end-points. The first endpoint is at x=2, and f(2)=a-2.